Let's solve each subpart step by step. --- # Given Data - **Initial Pressure** (\( p_1 \)): 2. bar (200 kPa) - **Quality** (\( x_1 \)): 65% = .65 - **Mass of H₂O** (\( m \)): 4 kg - **Rigid tank:** Volume is constant --- ## a. **Determine the volume of the tank, \( V_1 \), in m³.** ### Step 1: Find Specific Volume at State 1 From steam tables at \( p_1 = 2. \) bar (200 kPa): - \( v_f @ 200\, \text{kPa} \) (saturated liquid): ≈ .001061 m³/kg - \( v_g @ 200\, \text{kPa} \) (saturated vapor): ≈ .8857 m³/kg The mixture's specific volume: \[ v_1 = v_f + x_1 (v_g - v_f) \] \[ v_1 = .001061 + .65 \times (.8857 - .001061) \] \[ v_1 = .001061 + .65 \times .884639 \] \[ v_1 = .001061 + .574015 \] \[ v_1 \approx .5751\, \text{m}^3/\text{kg} \] ### Step 2: Calculate Total Volume \[ V_1 = m \cdot v_1 = 4\, \text{kg} \times .5751\, \text{m}^3/\text{kg} \] \[ V_1 \approx 2.30\, \text{m}^3 \] --- ## b. **Determine the total internal energy of the liquid-vapor mixture at state 1** From steam tables at \( p_1 = 2. \) bar: - \( u_f @ 200\, \text{kPa} \) ≈ 504.43 kJ/kg - \( u_g @ 200\, \text{kPa} \) ≈ 2506.5 kJ/kg Mixture's specific internal energy: \[ u_1 = u_f + x_1 (u_g - u_f) \] \[ u_1 = 504.43 + .65 \times (2506.5 - 504.43) \] \[ u_1 = 504.43 + .65 \times 2002.07 \] \[ u_1 = 504.43 + 1301.35 \] \[ u_1 \approx 1805.8\, \text{kJ/kg} \] Total internal energy: \[ U_1 = m \cdot u_1 = 4 \times 1805.8 = 7223.2\, \text{kJ} \] --- ## c. **Determine the pressure of state 2** **Tank is rigid:** Volume and mass constant, so specific volume \( v_2 = v_1 = .5751\, \text{m}^3/\text{kg} \). **When heat is added:** Water will eventually become all vapor (when \( x_2 = 1 \)), then superheated. To find pressure at state 2, look for the pressure at which the specific volume of saturated vapor equals \( v_1 \): From steam tables: - At \( p \approx .6\, \text{bar} \) (60 kPa): \( v_g = 3.240\, \text{m}^3/\text{kg} \) - At \( p \approx 2.\, \text{bar} \): \( v_g = .8857\, \text{m}^3/\text{kg} \) - At \( p \approx 3.\, \text{bar} \): \( v_g = .6058\, \text{m}^3/\text{kg} \) - At \( p \approx 3.5\, \text{bar} \): \( v_g = .5191\, \text{m}^3/\text{kg} \) \( .5751\, \text{m}^3/\text{kg} \) is between \( v_g \) at 3. and 3.5 bar. **Interpolate:** \[ P_2 = 3. + \frac{.6058-.5751}{.6058-.5191} \times (3.5-3.) \] \[ = 3. + \frac{.0307}{.0867} \times .5 \] \[ = 3. + .177 \approx 3.18\, \text{bar} \] --- ## d. **Determine the temperature of the water at state 2, \( T_2 \) in K** At state 2, all vapor (\( x_2 = 1 \)), at \( P_2 \approx 3.18 \) bar From steam tables, saturated temperature at 3.18 bar: - At 3. bar: \( T_{sat} = 133.6^\circ C \) - At 3.5 bar: \( T_{sat} = 141.9^\circ C \) Interpolate: \[ T_2 = 133.6 + \frac{3.18-3.}{3.5-3.} \times (141.9-133.6) \] \[ = 133.6 + .36 \times 8.3 = 133.6 + 2.988 \approx 136.6^\circ C \] Convert to Kelvin: \[ T_2 = 136.6 + 273.15 = 409.75\, \text{K} \] --- ## e. **Determine how much heat must be added to the tank to obtain state 2** Find internal energy at state 2 (\( x_2 = 1 \), \( P_2 = 3.18 \) bar): From steam tables: - At 3. bar: \( u_g = 2537.6\, \text{kJ/kg} \) - At 3.5 bar: \( u_g = 2543.7\, \text{kJ/kg} \) Interpolate: \[ u_2 = 2537.6 + \frac{3.18-3.}{3.5-3.} \times (2543.7-2537.6) \] \[ = 2537.6 + .36 \times 6.1 = 2537.6 + 2.196 \approx 2539.8\, \text{kJ/kg} \] Total internal energy at state 2: \[ U_2 = 4 \times 2539.8 = 10159.2\, \text{kJ} \] Heat added \( Q = U_2 - U_1 \): \[ Q = 10159.2 - 7223.2 = 2936\, \text{kJ} \] --- ## f. **The heating continues until the water in the tank reaches state 3 with a pressure \( p_3 = 5 \) bar. What is the temperature of the water in the tank at state 3, \( T_3 \)?** Volume is still \( V_1 = 2.30\, \text{m}^3 \), mass = 4 kg, so \( v_3 = v_1 = .5751\, \text{m}^3/\text{kg} \). At \( p_3 = 5 \) bar: - \( v_g = .3744\, \text{m}^3/\text{kg} \) (saturated vapor) - \( v_3 > v_g \), so the water is **superheated vapor**. Look up superheated steam tables for \( p = 5 \) bar, \( v = .5751\, \text{m}^3/\text{kg} \): - At \( T = 400^\circ C \), \( v = .5958\, \text{m}^3/\text{kg} \) - At \( T = 300^\circ C \), \( v = .4804\, \text{m}^3/\text{kg} \) Interpolate: \[ T_3 = 300 + \frac{.5751-.4804}{.5958-.4804} \times (400-300) \] \[ = 300 + \frac{.0947}{.1154} \times 100 \] \[ = 300 + .821 \times 100 = 300 + 82.1 \approx 382^\circ C \] Convert to Kelvin: \[ T_3 = 382 + 273.15 = 655.15\, \text{K} \] --- # **Summary Table** | Part | Quantity | Value | |------|---------------------------------------|-------------------| | a | Volume of tank, \( V_1 \) | \( 2.30\, \text{m}^3 \) | | b | Total internal energy, \( U_1 \) | \( 7223.2\, \text{kJ} \) | | c | Pressure at state 2, \( P_2 \) | \( 3.18\, \text{bar} \) | | d | Temperature at state 2, \( T_2 \) | \( 409.8\, \text{K} \) | | e | Heat added to reach state 2 | \( 2936\, \text{kJ} \) | | f | Temperature at state 3, \( T_3 \) | \( 655.2\, \text{K} \) | --- If you need a deeper explanation for any part or want to see the interpolation steps more explicitly, let me know!